Three different vegetables, turnips, eggplant and asparagus, make Bill sick sometimes. There is a 50% chance he will be sick if he eats turnips. If he eats eggplant, there is a 20% chance he will be sick. The probability that he will be sick when he eats asparagus is only 5%.Bill gets sick after dinner where he ate one of those three vegetables. What is the probability that he ate turnips? (Assume that for each of the three vegetables, there is a 1/3 chance he ate it.)

The probability that Bill ate turnips for dinner given that he gets sick is 0.6667.

Step-by-step explanation:

Denote the events as follows:

T = Bill eats turnips

E = Bill eats eggplant

A = Bill eats asparagus

X = Bill gets sick.

Given:

P (X | T) = 0.50

P (X | E) = 0.20

P (X | A) = 0.05

P (T) = P (E) = P (A) = 0.33

Compute the value of P (X) using the law of total probability as follows:

The probability that Bill gets sick is 0.2475.

Compute the probability that Bill ate turnips for dinner given that he gets sick as follows:

Thus, the probability that Bill ate turnips for dinner given that he gets sick is 0.6667.

the answer is 0.071

good luck! : d